91Ó°ÊÓ

The denominator of the function is \(x\), which is undefined at \(x=0\). So that's one critical point to consider.

For the numerator \(4x - 1\) to be equal to zero, \(x\) must be \(\frac{1}{4}\). This is our second critical point.

Our critical points divide the number line into three intervals: \(x < 0\), \(0 < x < \frac{1}{4}\), and \(x > \frac{1}{4}\). We plug a test point from each interval into the initial inequality to determine the sign of the values within that interval. For \(x < 0\), we can pick \(-1\), so the inequality \(\frac{4(-1)-1}{-1}\) is less than 0, which is false. That means \(x\) can't be less than 0. For \(0 < x < \frac{1}{4}\), we can pick \(\frac{1}{8}\), so the inequality \(\frac{4(\frac{1}{8})-1}{\frac{1}{8}}\) is greater than 0, which is true. That means \(x\) can be between 0 and \(\frac{1}{4}\). For \(x > \frac{1}{4}\), we can pick 1, so the inequality \(\frac{4(1)-1}{1}\) is greater than 0, which is also true. That means \(x\) can be greater than \(\frac{1}{4}\).

Now we can graph the solution set. We draw a number line and mark the critical points \(0\) and \(\frac{1}{4}\). For the interval between \(0\) and \(\frac{1}{4}\), which includes the solutions, we draw a line with an open circle at \(0\) and a closed circle at \(\frac{1}{4}\) since \(0\) is excluded and \(\frac{1}{4}\) is included. For the interval greater than \(\frac{1}{4}\), we draw a ray beginning at \(\frac{1}{4}\) and extending to the right.